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Appendix for The Cannonball Approach to Space Travel
Appendix for The Cannonball Approach to Space
Travel
Morten Monrad Pedersen
September 25, 2003
1
Introduction
This appendix contains some computations that for reasons of readability were left out
of the article The Cannonball Approach to Space Travel. If you want to study the
physics behind these computation I refer you to General Physics by Sternheim and
Kane or any other physics textbook dealing with mechanics.
I start out by showing how to compute the throwing distances for objects thrown
at an angle of 45 degrees. I do this to show that these standard computations give the
same results that Peeters arrives at, and that it is therefore very likely that it is exactly
these computations that Peeters has used. This should establish that his argument has
throwing computations suers from the bad assumptions that I mention in the main
article.
After that I go on to compute the correct escape velocity for Earth, and nally
I compute the orbital velocities of the International Space Station and satellites in
geosynchronous orbit (geosynchronous orbit is the orbit in which satellites stay in the
same position all the time relative to the surface of Earth).
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Throwing Distances
In this section I will look at throwing distances for objects thrown at an angle of 45
degrees, with the assumption that there is no air resistance. It is furthermore assumed
that the area of the Earth that we look at is at, and that the gravitational acceleration
g is always equal to 10
m
s
(this is the value used by Peeters).
Let v
i
denote the initial velocity of the thrown object, let v
x
be the velocity in the
horizontal direction, and let v
y
represent the initial velocity in the vertical direction.
We can compute v
x
and v
y
from v
v
x
= v
y
= v
i
sin 45 .
(1)
The time of ight t is the time it takes for the vertical speed to be reduced to zero
and then increased to the initial value v
y
again (see gure 1). This means that we can
nd t
1 Vertical speed = v
y
Vertical speed = v
y
Vertical speed = 0
Figure 1: A thrown object will gain height until gravity has brought its vertical velocity
down to zero. After that the ball will fall towards the Earth, and since we ignore air
resistance it will hit Earth with a velocity equal to the initial speed (but in the opposite
direction)
2v
y
= gt
(2)
t = 2v
y
g
(3)
Since we assume that there is no air resistance, the horizontal distance traveled d
x
is simply the time of ight times the horizontal velocity
d
x
= v
x
t
(4)
= v
x
2v
y
g
, by equation (3)
(5)
= 2v
2
i
sin
2
45 g
, by equation (1)
(6)
= v
2
i
g
, since sin 45 = 1
2.
(7)
Peeters computes the throwing distances for at number of initial velocities. With an
initial velocity of 8000
m
s
he gets a throwing distance of 6400 km. I will now compute
the throwing distance for an initial velocity of 8000
m
s
using equation (7).
d
x
= 8000
m
s
2
10
m
s
2
(8)
= 6400 km.
(9)
This is exactly the same value that Peeters presents, but the computation is, as I
have argued in the main article, based on some rather dubious assumptions.
2 3
Escape Velocity
In this section I will compute the correct escape velocity for Earth.
The potential energy E
p
of an object in Earths gravitational eld is
E
p
= GM m
r
,
(10)
where G = 6.67 · 10
11 Nm
2
kg
2
is the gravitational constant, M = 5.98 · 10
24
kg is the mass
of Earth, m is the mass of the object, and r is the distance of the object from the center
of the Earth. Note that the reference level (E
p
= 0) is chosen as an innitely distant
position.
The kinetic energy E
k
of an object is
E
k
= 1
2 mv
2
,
(11)
where v is the velocity of the object.
If an projectile is to be red away from Earth, such that it will never be pulled back,
the kinetic energy of the object must at least as large as the potential energy it will gain.
Therefore we can compute the escape velocity v
e
based on equations (10) and (11). Let
R = 6.38 · 10
6
m denote the radius of the Earth, let E
k,surf ace
denote the kinetic energy
of the object, when it is at the surface of Earth, let E
p,surf ace
denote the potential energy
of an object at the surface of Earth and recall that the potential energy will be zero
when the object has escaped Earths gravity.
E
k,surf ace
= 0 E
p,surf ace
(12)
1
2 mv
2
e
= GM m
R
(13) v
e
=
2GM
R
,
(14)
Equation (14) enables us to arrive at the escape velocity for Earth
v
e 2 · 6.67 · 10
11 Nm
2
kg
2
· 5.98 · 10
24
kg
6.38 · 10
6
m
(15)
11200 m
s
(16)
40300 km
h
(17)
4
The International Space Station and Satellites in
Geosynchronous Orbit
In this nal section I will compute the orbital velocities of the International Space
Station and satellites in geosynchronous orbit.
The gravitational force F
g
from Earth on an object a distance r from the center of
Earth is
3 F
g
= GM m
r
2
.
(18)
The centrifugal force F
c
on an object with mass m and velocity v in an orbit around
Earth with a radius of r is
F
c
= mv
2
r .
(19)
If an object is to remain in circular orbit around the Earth the gravitational force
exerted on it by Earth must be equal to the centrifugal force on it due to its circular
movement.
Therefore we can use equations (18) and (19) nd the orbital velocity v
o
of an object
with mass m in a circular orbit with a radius of r
F
g
= F
c
(20)
GMm
r
2
= mv
2
o
r
(21) v
o
=
GM
r
(22)
According to http://www.wikipedia.org/wiki/International Space Station the orbit
radius of the International Space Station is on average 386 km (the actual height varies
over time by several kilometres due to atmospheric drag and reboosts). This means
that the average distance from the center of the Earth is roughly 6770 km.
Armed with equation (22) and the orbit radius we can compute the orbital speed
v
ISS
of the International Space Station
v
ISS 6.67 · 10
11 Nm
2
kg
2
· 5.98 · 10
24
kg
6.77 · 10
6
m
(23)
7680 m
s
(24)
27600 km
h
(25)
According to http://www.wikipedia.org/wiki/Geosynchronous orbit the geosynchronous
orbit has a radius of 42163 km, using this and equation (22) we can compute the orbital
velocity v
gso
of satellites in geosynchronous orbit
v
gso 6.67 · 10
11 Nm
2
kg
2
· 5.98 · 10
24
kg
42.2 · 10
6
m
(26)
3070 m
s
(27)
11100 km
h
(28)
It is seen that the orbital velocities of both the International Space Station and
satellites in geosynchronous orbit are larger than the 4000
km
h
upper speed limit claimed
by Peeters for objects with a mass of 1 kg or larger.
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